Compactness of Cofinite and Indiscrete Spaces

Two Topologies with Very Different Open Sets

Compactness asks whether every open cover contains a finite subcover. In some spaces, the open sets have a form that makes this question particularly direct: a cofinite open set omits only finitely many points, while an indiscrete space has just one possible nonempty open set. These descriptions are enough to prove compactness without any metric or separation assumptions.

An open cover of \(X\) is a family \(\mathcal{U}\) of open subsets of \(X\) whose union is \(X\). A finite subcover is a finite subfamily of \(\mathcal{U}\) whose union is still \(X\). A space is compact if every open cover has a finite subcover.

In the cofinite definition, \(X\) itself is open because its complement is empty, which is finite. When \(X\) is finite, every subset has finite complement, so the cofinite topology is the discrete topology. When \(X\) is infinite, it is generally much coarser: its open sets are the empty set and the sets that leave out only finitely many points.

The displayed collections are indeed topologies. For the cofinite collection, \(\varnothing\) and \(X\) belong to it. The complement of a union of any collection of cofinite open sets is contained in the finite complement of any one nonempty member of that collection, so that union has finite complement; if all members are empty, the union is empty. For a finite intersection of cofinite open sets, its complement is a finite union of finite sets, hence finite. The empty set cases also satisfy the topology axioms. The indiscrete collection contains \(\varnothing\) and \(X\); any union or finite intersection of its members is again either \(\varnothing\) or \(X\).

Compactness of Cofinite Spaces

Proof. Let \(\mathcal{U}\) be an open cover of \(X\). If \(X=\varnothing\), the empty subfamily is a finite subcover, since its union is \(\varnothing=X\). Now suppose \(X\ne\varnothing\). Since \(\mathcal{U}\) covers \(X\), at least one member \(U_0\in\mathcal{U}\) is nonempty. By the definition of the cofinite topology, \(F=X\setminus U_0\) is finite. If \(F=\varnothing\), then \(U_0=X\), so \(\{U_0\}\) is a finite subcover. If \(F\ne\varnothing\), for each \(x\in F\) choose a cover member \(U_x\in\mathcal{U}\) with \(x\in U_x\); such a member exists because \(\mathcal{U}\) covers \(X\). There are only finitely many points in \(F\), so these choices produce only finitely many cover members. The family \(\{U_0\}\cup\{U_x:x\in F\}\) covers \(U_0\) and every point of \(F\), and \(X=U_0\cup F\). It is therefore a finite subcover. Only finitely many choices were needed, and no choice principle for arbitrary families is required. \(\square\)

The central step is to start with one open set and inspect only the points it misses. Cofiniteness says that there are finitely many such points; the cover supplies one additional member for each. The argument applies whether \(X\) is finite or infinite, and it also handles the case where the first open set already equals \(X\).

Compactness of Indiscrete Spaces

Proof. Let \(\mathcal{U}\) be an open cover of \(X\). If \(X=\varnothing\), then the empty subfamily covers \(X\). If \(X\ne\varnothing\), the cover must contain at least one nonempty member: otherwise its union would be empty and could not equal \(X\). The only nonempty open set in the indiscrete topology is \(X\) itself. Therefore some member of \(\mathcal{U}\) equals \(X\), and that single member is a finite subcover. \(\square\)

Here the cover’s defining condition does nearly all the work. A cover of a nonempty indiscrete space cannot cover any point with a proper nonempty open set, because there are no such open sets. It must include \(X\) itself.

A Common Mistake and a Separation Reminder

Common pitfall. An infinite open cover need not lack a finite subcover. In the cofinite topology on \(\mathbb{N}\), the family \(\{U_n:n\in\mathbb{N}\}\), where \(U_n=\mathbb{N}\setminus\{n\}\), has infinitely many members. A mistaken argument might say, “There are infinitely many sets, so no finite subcover exists.” But the computation \(U_1\cup U_2=(\mathbb{N}\setminus\{1\})\cup(\mathbb{N}\setminus\{2\})=\mathbb{N}\) shows that two members do cover. The number of sets in the original cover is not the test; the test is whether some finite selection still has union \(X\).

Compactness also does not imply that a space is Hausdorff. For example, take an infinite set \(X\) with the cofinite topology, and let \(x,y\in X\) be distinct. If \(U,V\) are nonempty open neighborhoods of \(x,y\), respectively, then \(X\setminus U\) and \(X\setminus V\) are finite. Their union is finite, and \(U\cap V=X\setminus\big((X\setminus U)\cup(X\setminus V)\big)\). Since \(X\) is infinite, removing a finite set cannot leave the empty set, so \(U\cap V\ne\varnothing\). Thus \(x\) and \(y\) have no disjoint open neighborhoods. In the indiscrete topology on a set with at least two points, every open neighborhood of any point is \(X\), so distinct points also cannot have disjoint neighborhoods. Both kinds of spaces are compact, but these examples show that compactness alone does not supply Hausdorff separation.

The two compactness proofs use different features of the open sets. Cofinite open sets leave a finite remainder to be covered; an indiscrete cover of a nonempty space must contain the entire space as one member. In either case, the conclusion follows directly from the finite-subcover definition, with no assumption that the space is Hausdorff.